Question

The passage of current liberates `H_2` at cathode and `Cl_2` at anode. The solution is
A. copper chloride in water
B. NaCl in water
C. `H_2SO_4`
D. Water

Answer 1

Correct Answer - B
Since discharge potential of watier is greater than that of sodium so water is reduced at cathode instead of `Na^(+)`.
Cathode: `H_(2)O+e^(-)to(1)/(2)H_(2)+OH^(-)`
Anode: `Cl^(-)to(1)/(2)Cl_(2)+e^(-)`.